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Assuming Complete Dissociation, Calculate the Ph of the Following Solutions: 0.003 M Hcl - Chemistry

Assuming complete dissociation, calculate the pH of the following solutions:

0.003 M HCl

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Solution

0.003MHCl

`H_2O + HCl ↔  H_3O^+   + Cl^(-)` 

Since HCl is completely ionized,

`[H_3O^+] = [HCl]`

`=>[H_3O^+] = 0.003`

Now

`pH = - log[H_3O^+] = -log(.003)`

= 2.52

Hence, the pH of the solution is 2.52.

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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 7 Equilibrium
Q 48.1 | Page 229
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